Question

Is there way to initialize a numpy array of a shape and add to it? I will explain what I need with a list example. If I want to create a list of objects generated in a loop, I can do:

a = []
for i in range(5):
    a.append(i)

I want to do something similar with a numpy array. I know about vstack, concatenate etc. However, it seems these require two numpy arrays as inputs. What I need is:

big_array # Initially empty. This is where I don't know what to specify
for i in range(5):
    array i of shape = (2,4) created.
    add to big_array

The big_array should have a shape (10,4). How to do this?

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EDIT:

I want to add the following clarification. I am aware that I can define big_array = numpy.zeros((10,4)) and then fill it up. However, this requires specifying the size of big_array in advance. I know the size in this case, but what if I do not? When we use the .append function for extending the list in python, we don’t need to know its final size in advance. I am wondering if something similar exists for creating a bigger array from smaller arrays, starting with an empty array.

Answer 1

numpy.zeros

Return a new array of given shape and
type, filled with zeros.

or

numpy.ones

Return a new array of given shape and
type, filled with ones.

or

numpy.empty

Return a new array of given shape and
type, without initializing entries.

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However, the mentality in which we construct an array by appending elements to a list is not much used in numpy, because it’s less efficient (numpy datatypes are much closer to the underlying C arrays). Instead, you should preallocate the array to the size that you need it to be, and then fill in the rows. You can use numpy.append if you must, though.